exponential equation with different bases We have $3^x-5^\frac{x}{2}=4$  My question is what we can do here ? Can we solved it algebraically or we need to notice that $x=2$ and then show that for $x \neq 2$ there aren't any other solutions? 
 A: The given equation holds for the roots of the function
$$f(x) = 3^x-\sqrt{5}^x-4$$
The derivate is
$$f '(x)=3^xln{3}-\sqrt{5}^xln{(\sqrt{5}})$$
So, f '(x) > 0 for all x > 0.
So, f(x) has at most one root.
This root can only be calculated with numerical methods, but in this case,
it is easy to guess the solution.
A: Your second approach, viz. showing there are no other solutions would also require some algebra.  With $2t = x$, you can write the equation as
$$(4+5)^t = 4 + 5^t$$
This is obvious for $t=1$ i.e $x=2$.
For $t > 1$, we have  $(4+5)^t > 4^t+5^t > 4+5^t$
for $0 < t < 1$, let $y = \frac1t > 1$, then $(4+5^t)^y > 4^y+5 > 4+5 = \left((4+5)^t\right)^y$
and for $t \le 0$, $(4+5)^t \le 1 < 4 < 4+5^t $ 
A: To go one step further, let me consider the equation $3^x-5^\frac{x}{2}=a$ where $a$ is a positive constant.  
What Peter showed is that, since the derivative is almays positive, the equation has only one root. In tour case, the problem was simple since, for $a=4$, there is an obvious solution at $x=2$.  
Where the problem starts to be different is when there is not obvious solution; except using inspection to try to find two values of $x$ which bracket the solution, the only other solution is to plot the fuction for the same goal. This wille give you a rough estimate I shall write $x_\text{old}$; this value will be the starting point of a root-finder method. Here, for simplicity, I shall use Newton method.
So, for an example, let me chose $a=12345.6789$. Plotting the function shows that the solution is somwhere between $8$ and $9$. So, being lazy, I shall start the iterations at $x_\text{old} = 8$.  
As you know, Newton iteration scheme write  
$x_\text{new} = x_\text{old} - f(x_\text{old}) / f'(x_\text{old})$  
and, after each iteration, $x_\text{old}$ is updated by $x_\text{new}$.  
So, for the case I selected, the following iterates will appear : $8.00000$, $8.95595$,  $8.69676$, $8.65122$, $8.65003$ and this is the end of the process for six significant figures. For sure, you can continue iterating until you reach the desired level of accuracy.
